# Indefinite integral…for calc 1?

a calc 1 student I know was given this integral do via u-substitution, and even though I think it's an obvious typo, I know the answer is somewhat simple thanks to Wolfram, but am not sure how to arrive at it.

$\int{x^2\sqrt{1+x}}$ $dx$.

Any help would be appreciated!

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## 1 Answer

let $u=1+x$ so that $du=dx$ and $x^2=(u-1)^2$

The integral now becomes $\int(u-1)^2(\sqrt{u}) du=\int(u^{5/2}-2u^{3/2}+u^{1/2})du=\frac{2}{7}u^{7/2}-\frac{4}{5}u^{5/2}+\frac{2}{3}u^{3/2}+c$

Remember to replace $u$ by $x+1$ in the last expression.

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Derp...6 years will do that to you! Thanks! –  Johnny Apple Nov 15 '13 at 4:11
Factoring out $u^{\frac{3}{2}}$ from the result makes the re-substitution a lot easier... –  User58220 Nov 15 '13 at 4:35